Grouping an array by spreading another array

Question

Previously, structure of my array was:

[
  {
    "id": "Ddfwe23224533",
    "kind": "PEAK_EXPERIENCE",
    "title": "IP - Collective Identity",
  },
  {
    "id": "Ddfwe23224534",
    "kind": "PEAK_EXPERIENCE",
    "title": "IP - Collective Identity 2",
  },
  .....
]

I have created a function called flatToHierarchy to group the elements in an array. Here is the code:

export const flatToHierarchy = (arr, prop) => arr.reduce((accumulator, currentItem) => {
  const items = accumulator;
  const propValue = currentItem[prop];
  items[propValue] = items[propValue]
    ? [...items[propValue], currentItem]
    : [currentItem];
  return items;
}, {});

Now if I call it like:

flatToHierarchy(elements, 'kind')

It works fine.

Now, structure of my array is changed. Now it looks like:

[
  {
    "entity": {
      "id": "Ddfwe23224533",
      "kind": "PEAK_EXPERIENCE",
      "title": "IP - Collective Identity",
    },
    entity_id: "jj98834jfj983"
  },
  {
    "entity": {
      "id": "Ddfwe23224534",
      "kind": "PEAK_EXPERIENCE",
      "title": "IP - Collective Identity 2",
    },
    entity_id: "jj98834jfdf83"
  },
  .....
]

So, now for grouping array by element kind, I call it this way:

flatToHierarchy(elements, 'entity.kind')

Now, it yields me something like:

{
  undefined: [....element1, ....element2 ]
}

But what I want is:

{
  PEAK_EXPERIENCE: [....element1, ....element2 ]
}

I have dug deeper in to this and found that array cannot be accessed this way:

array['something.something']

It needs to be accessed this way:

array['something']['something']

So, I changed my function but I am stuck. Here is my changed function:

export const flatToHierarchy = (arr, prop) => arr.reduce((accumulator, currentItem) => {
  const items = accumulator;
  const props = prop.split('.').map(p => [p]);
  console.log(props);
  const propValue = currentItem[prop];
  console.log(currentItem...props); // how to spread the array here?
  items[propValue] = items[propValue]
    ? [...items[propValue], currentItem]
    : [currentItem];
  return items;
}, {});

Update:

Sample of current input:

[
  {
    "entity": {
      "id": "Ddfwe23224533",
      "kind": "PEAK_EXPERIENCE",
      "title": "IP - Collective Identity",
    },
    entity_id: "jj98834jfj983"
  },
  {
    "entity": {
      "id": "Ddfwe23224534",
      "kind": "PEAK_EXPERIENCE",
      "title": "IP - Collective Identity 2",
    },
    entity_id: "jj98834jfdf83"
  },
  {
    "entity": {
      "id": "Ddfwe23224594",
      "kind": "PEAK_EXPERIENCE2",
      "title": "IP - Collective Identity 6",
    },
    entity_id: "jj98834jfdf33"
  },
]

sample of needed output:

{
  PEAK_EXPERIENCE: [
    {
      "entity": {
        "id": "Ddfwe23224533",
        "kind": "PEAK_EXPERIENCE",
        "title": "IP - Collective Identity",
      },
      entity_id: "jj98834jfj983"
    },
    {
      "entity": {
        "id": "Ddfwe23224534",
        "kind": "PEAK_EXPERIENCE",
        "title": "IP - Collective Identity 2",
      },
      entity_id: "jj98834jfdf83"
    },
  ],
  PEAK_EXPERIENCE2: [
    {
      "entity": {
        "id": "Ddfwe23224594",
        "kind": "PEAK_EXPERIENCE2",
        "title": "IP - Collective Identity 6",
      },
      entity_id: "jj98834jfdf33"
    },
  ],
}

Update 2:

While making changes to the required function, I need to keep in mind that this function will always have the signature as specified below:

flatToHierarchy(array, string);

example:

flatToHierarchy(elements, 'kind');

another example:

flatToHierarchy(elements, 'entity.kind');

This question is a bit unclear. Could you please post a sample of the input (which i think is there) and one of final output that you're looking for. — rmn
– rmn, Commented Sep 7, 2018 at 11:10
@rmn I have included the samples in update part of the question. Please have a look at them. — Vishal
– Vishal, Commented Sep 7, 2018 at 11:18

user3297291 · Accepted Answer · 2018-09-07 11:29:35Z

4

You're looking for a path function, which you can make by using reduce and seeding with the initial object.

Watch out for null/undefined properties!

const propValue = prop
  .split('.') // Make a list of all property names (cannot contain a .)
  .reduce((obj, p) => obj[p], currentItem); // Loop over them to traverse your object

const flatToHierarchy = (arr, prop) => arr.reduce((accumulator, currentItem) => {
  const items = accumulator;
  const propValue = prop.split('.').reduce((obj, p) => obj[p], currentItem);
  items[propValue] = items[propValue]
    ? [...items[propValue], currentItem]
    : [currentItem];
  return items;
}, {});

const data = [
  {
    "entity": {
      "id": "Ddfwe23224533",
      "kind": "PEAK_EXPERIENCE",
      "title": "IP - Collective Identity",
    },
    entity_id: "jj98834jfj983"
  },
  {
    "entity": {
      "id": "Ddfwe23224534",
      "kind": "PEAK_EXPERIENCE",
      "title": "IP - Collective Identity 2",
    },
    entity_id: "jj98834jfdf83"
  },
  {
    "entity": {
      "id": "Ddfwe23224594",
      "kind": "PEAK_EXPERIENCE2",
      "title": "IP - Collective Identity 6",
    },
    entity_id: "jj98834jfdf33"
  },
]

console.log(
  flatToHierarchy(data, "entity.kind")
);

answered Sep 7, 2018 at 11:29

user3297291

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1 Comment

Vishal Over a year ago

Thanks for the answer. This is what I wanted.

rmn · Accepted Answer · 2018-09-07 11:23:32Z

0

Considering your sample input is in a variable called data, you can do the following.

data = data.reduce((output, current) => {
    const kind = current.entity.kind
    output[kind] = (output[kind] || []).concat(current)
    return output
}, {})

answered Sep 7, 2018 at 11:23

rmn

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4 Comments

Vishal Over a year ago

thanks for the answer. But entity.kind will not be known. It will be passed to function call. Look at the update 2 part of the question.

rmn Over a year ago

Got it. What happens when the string is just 'kind' as in your first example. Since, there's no 'kind', just 'entity.kind', will it fail?

Vishal Over a year ago

it will return an empty array.

rmn Over a year ago

For anyone referencing it in future, since the question was updated, please see the correct answer above.

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